Solve the equation `x^(2)-(sqrt(3)+1)x+sqrt(3)=0` by the method of completing the square.

Solve the equation `x^(2)-(sqrt(3)+1)x+sqrt(3)=0` by the method of com

1.	Solve the equation `x^(2)-(sqrt(3)+1)x+sqrt(3)=0` by the method of completing the square.
Answer» We have `x^(2)-(sqrt(3)+1)x+sqrt(3)=0` `implies" "x^(2)-(sqrt(3)+1)x=-sqrt(3)` `implies" "x^(2)-2xx x xx((sqrt(3)+1)/(2))+((sqrt(3)+1)/(2))^(2)=-sqrt(3)+((sqrt(3)+1)/(2))^(2)` `" ""[""adding "((sqrt(3)+1)/(2))^(2)" on both sides""]"` `implies" "{x-((sqrt(3)+1))/(2)}^(2)={((sqrt(3)+1)^(2))/(4)-sqrt(3)}` `implies" "("("sqrt(3)+1")"^(2)-4sqrt(3))/(4)=((sqrt(3)-1)/(2))^(2)` `implies" "{x-("("sqrt(3)+1")")/(2)}=+-("("sqrt(3)-1")")/(2)" "["taking square root on both sides"]` `implies" "x-("("sqrt(3)+1")")/(2)=("("sqrt(3)-1")")/(2)" or "x-("("sqrt(3)+1")")/(2)=(-"("sqrt(3)-1")")/(2)` `implies" "x=("("sqrt(3)-1")")/(2)+("("sqrt(3)+1")")/(2)=(2sqrt(3))/(2)=sqrt(3)` or `x=("("-sqrt(3)+1)/(2)+("("sqrt(3)+1")")/(2)=(2)/(2)=1` `implies" "x=sqrt(3)" or "x=1.` Hence, `sqrt(3)` and 1 are the roots of the given equation.